Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $a = \dfrac{-4q + 28}{q + 2} \times \dfrac{q^3 + 5q^2 + 6q}{-q^3 - 5q^2 - 6q} $
First factor out any common factors. $a = \dfrac{-4(q - 7)}{q + 2} \times \dfrac{q(q^2 + 5q + 6)}{-q(q^2 + 5q + 6)} $ Then factor the quadratic expressions. $a = \dfrac {-4(q - 7)} {q + 2} \times \dfrac {q(q + 3)(q + 2)} {-q(q + 3)(q + 2)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac {-4(q - 7) \times q(q + 3)(q + 2) } {(q + 2) \times -q(q + 3)(q + 2) } $ $a = \dfrac {-4q(q + 3)(q + 2)(q - 7)} {-q(q + 3)(q + 2)(q + 2)} $ Notice that $(q + 3)$ and $(q + 2)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac {-4q\cancel{(q + 3)}(q + 2)(q - 7)} {-q\cancel{(q + 3)}(q + 2)(q + 2)} $ We are dividing by $q + 3$ , so $q + 3 \neq 0$ Therefore, $q \neq -3$ $a = \dfrac {-4q\cancel{(q + 3)}\cancel{(q + 2)}(q - 7)} {-q\cancel{(q + 3)}(q + 2)\cancel{(q + 2)}} $ We are dividing by $q + 2$ , so $q + 2 \neq 0$ Therefore, $q \neq -2$ $a = \dfrac {-4q(q - 7)} {-q(q + 2)} $ $ a = \dfrac{4(q - 7)}{q + 2}; q \neq -3; q \neq -2 $